Practice question
Question
How much heat is required to raise 0.6kg of water from 25∘C to 75∘C and then convert 0.4kg of it to steam at 100∘C in a 0.1kg copper calorimeter initially at 25∘C? (Specific heat of water = 4186J kg−1K−1, copper = 386J kg−1K−1, latent heat of vaporization = 2.256×106J kg−1)
Explanation
Q1 = (0.6×4186+0.1×386)×(75−25) = (2511.6+38.6)×50 = 2550.2×50 = 127510J (to 75°C). Q2 = (0.6×4186+0.1×386)×(100−75) = 2550.2×25 = 63755J (to 100°C). Q3 = 0.4×2.256×106 = 902400J (vaporization). Total: Q = 127510+63755+902400 = 1093665J = 1093.67kJ.